Question: Find $\lim_{x\to 3}\dfrac{9-x^2}{\cos\left(\dfrac\pi2x\right)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $9$ (Choice B) B $\pi$ (Choice C) C $-\dfrac{12}{\pi}$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=3$ into $\dfrac{9-x^2}{\cos\left(\dfrac\pi2x\right)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 3}\dfrac{9-x^2}{\cos\left(\dfrac\pi2x\right)} \\\\ &=\lim_{x\to 3}\dfrac{\dfrac{d}{dx}[9-x^2]}{\dfrac{d}{dx}\left[\cos\left(\dfrac\pi2x\right)\right]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 3}\dfrac{-2x}{-\dfrac\pi2\sin\left(\dfrac\pi2x\right)} \\\\ &=\dfrac{-2(3)}{-\dfrac\pi2\sin\left(\dfrac{3\pi}2\right)} \gray{\text{Substitution}} \\\\ &=-\dfrac{12}{\pi} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 3}\dfrac{\dfrac{d}{dx}[9-x^2]}{\dfrac{d}{dx}\left[\cos\left(\dfrac\pi2x\right)\right]}$ actually exists. In conclusion, $\lim_{x\to 3}\dfrac{9-x^2}{\cos\left(\dfrac\pi2x\right)}=-\dfrac{12}{\pi}$.